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Let H = l2 and (en) be the standard orthonormal basis for H ( note that for all x Î H, x = ån = 1¥ < x,en > en). Set Tn = e1[`(Ä)] en. Then Tn* = en[`(Ä)] e1
(i) limn ||Tnx || = limn || < x,en > e1 || = limn | < x,en > | = 0. So Tn ® 0 in the strong operator topology. But limn||Tn*e1 || = limn ||en || = 1, hence
Tn* dosen't converge to zero in the strong operator topology. So T ® T* is not continuous in the strong operator topology.
(ii) The involution is continuous with respect to the weak operator topology (since | < Tx,y > | = | < T*y,x > |). Hence (i) implies that the weak operator topology and the strong operator topology don't coincide on B(H).
(iii) ||Tn || = ||e1 ||||en || = 1, and by (i) Tn ® 0 in the strong operator topology. Therefore the operator norm is not continuous on B(H).
(iv) Let D = {n[1/2]Tn ; n Î N}. For each neighborhood U(0,x1, ¼,xm,e) of 0 in the strong operator topology, with xk = ån = 1¥ aknen, ||n[1/2]Tnxk|| = n[1/2]|akn|.
But for every 1 £ k £ m, ån = 1¥ |akn |2 < ¥, hence for each e there exists a natural number n such that n[1/2]|akn| < e. So that 0 belongs to the strong closure of D.
It follows from the principle of uniform boundedness and ||Ön Tn || = Ön that any sequence in D doesn't converge to 0 in the strong operator topology. Hence the strong operator is not metrizable. Similarly one can show that the weak operator topology is not metrizable.
(v) Let L be the set of all (n,U) in which n Î N and U is a neighborhood of 0 in the strong operator topology on B(H). Then L with the following relation is a directed set:
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